将随机键和值的列表转换为字典

Question

Suppose keys are represents by k _n and values are represented by v _n .

I have a list of keys and values like my_list_1 = [k1,v1,v2,v3,k2,v4,v5,k3,v6,v7,v8,v9,k4,v10]

Keys can be in repeated too in the list like my_list_2 = [k1,v1,v2,v3,k2,v4,v5,k3,v6,v7,v8,v9,k2,v10]

All the values which are followed by a particular key belongs to that key. For instance in my_list_1; v1,v2,v3 belongs to k1; v4,v5 belongs to k2; v6,v7,v8,v9 belongs to k3 and v10 belongs to k4. Therefore final dictionary would look like-

{
   k1: [v1,v2,v3] ,
   k2: [v4,v5] ,
   k3: [v6,v7,v8,v9],
   k4: [v10]
}

Similarly in case of my_list_2 it would be:

{
   k1: [v1,v2,v3] ,
   k2: [v4,v5,v10] ,
   k3: [v6,v7,v8,v9]
}

How can I convert this kind of list in the required dictionary?

Note: I already have functions to identify whether a particular item in list is a key or a value. Let's call these functions as isKey() and isValue().

isKey() returns True if an item is a key else returns False

isValue() returns True if an item is a value else returns False

Answer 1

Maybe you could consider using a defaultdict :

from collections import defaultdict
from pprint import PrettyPrinter

def is_key(s: str) -> bool:
    return s.startswith('k')

def is_value(s: str) -> bool:
    return s.startswith('v')

def convert_to_dict(my_list: list) -> dict:
    my_defaultdict = defaultdict(list)
    curr_key = None
    for s in my_list:
        if is_key(s):
            curr_key = s
        elif is_value(s):
            if curr_key is not None:
                my_defaultdict[curr_key].append(s)
    return dict(my_defaultdict)

my_list_1 = ['k1', 'v1', 'v2', 'v3', 'k2', 'v4', 'v5', 'k3', 'v6', 'v7', 'v8', 'v9', 'k4', 'v10']
my_dict_1 = convert_to_dict(my_list_1)
print("my_dict_1:")
PrettyPrinter().pprint(my_dict_1)

my_list_2 = ['k1', 'v1', 'v2', 'v3', 'k2', 'v4', 'v5', 'k3', 'v6', 'v7', 'v8', 'v9', 'k2', 'v10']
my_dict_2 = convert_to_dict(my_list_2)
print("\nmy_dict_2:")
PrettyPrinter().pprint(my_dict_2)

Output:

my_dict_1:
{'k1': ['v1', 'v2', 'v3'],
 'k2': ['v4', 'v5'],
 'k3': ['v6', 'v7', 'v8', 'v9'],
 'k4': ['v10']}

my_dict_2:
{'k1': ['v1', 'v2', 'v3'],
 'k2': ['v4', 'v5', 'v10'],
 'k3': ['v6', 'v7', 'v8', 'v9']}

Answer 2

Not sure if this can be done in a more pythonic way, but here is a loop that ought to do it. This assumes that a key will always directly precede all of its items and all of its items precede the next key.

def my_list_to_dict(my_list):
    my_dict = {}
    my_key = None
    my_values = []
    for item in my_list:
        if isKey(item):
            if my_key != None:
                my_dict[my_key] = my_values
            my_key = item
            my_values = []
        elif isValue(item):
            my_values.append(item)
    return my_dict

Answer 3

You could create a function like this one:

def to_dict(items):
    dictionary = {}
    current_key = None

    for item in items:
        if isKey(item):
            current_key = item
        elif current_key:
            if current_key in dictionary:
                dictionary[current_key].append(item)
            else:
                dictionary[current_key] = [item]

    return dictionary

And then call it like this:

dictionary_1 = to_dict(my_list_1)
dictionary_2 = to_dict(my_list_2)

Answer 4

You can use a dictionary and add each element to the dictionary using a loop while keeping track of the current key.

def to_dictionary(my_list):
    output = {}
    current_key = None
    for item in my_list:
        if isKey(item):
            current_key = item
        elif isValue(item):
            output[current_key] = output.get(current_key, []) + [item]
    return output

print(to_dictionary(my_list_1))

{'k1': ['v1', 'v2', 'v3'],
 'k2': ['v4', 'v5'],
 'k3': ['v6', 'v7', 'v8', 'v9'],
 'k4': ['v10']}